In another recent thread, a discussion raged about flow numbers for different turbos from various manufacturers and CFM/HP/lbs per min ratings. I posted a link I found a while back to a Turbo Buick site that had extensive mathematical formulas and information on air flow through an engine, volumetric efficiency, compressor maps, etc.

I apologize to those who already know this stuff, but for me it was a great learning experience, so I wanted to share. First, here are the formulas, where P is absolute pressure and T is temperature in degrees Rankine (Fahrenheit + 460):

To get pounds of air (lbs/min) when you already know the CFM:
P(psia) x V(cu.ft./min) x 29
(10.73 x T(deg R))

To get the volume of air (CFM) when you already know the lbs/min:
n(lbs/min) x 10.73 x T(deg R)
(29 x P(psia))

Someone had asked a question on converting a 36 lbs/min flow rate to CFM and after doing the calculations, the numbers I arrived at didn't look right. I e-mailed the author of the page, John Estill (a very nice and helpful guy, BTW) and he set me straight:

If you are trying to convert a lb/min number from a compressor map to cfm, then you are using nontypical values, which is why the number is lower than expected. 15 psi gauge, or 15 psi boost = 29.7 psia, which is not what you have at the compressor inlet! Turbonetics maps are done assuming a standard inlet pressure of 13.949 psia (about 1.5" Hg vacuum) and 85F. At these conditions your 36 lb/min of air flow = 520 cfm.

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So, in summary, when using these formulas to do calculations on lbs/min and CFM, use 13.949 for psia, and use 545 (which is 85 degrees F + 460 to get degrees Rankine, or deg R, in the formula) for the temp value. I don't know if I can answer detailed questions about this info, but I'd be happy to try. You can always e-mail John Estill about it, too-- his e-mail link is at the top of his page (the one I provided the link for above).

ooops

Thanks a lot doug you are very helpful.
and it was my question to begin with but I'm glad that somebody else was interested enough to find the answer.

I'm still trying to get rid of my headache I got from trying to get that formula down.

Question is how do you convert the 13.949 at a different temperature. I doubt its as easy say 50 degrees outside hence 85x=(19.949*50). Also, is that 13.949 always unchanged despite the amount of boost your running or does that need converted as well.

From what the guy who wrote the article, said, 13.949 psi and 85 degrees are just the "standard" numbers used in Turbonetics pressure maps. I don't know if other manufacturers also use those same values when developing their compressor maps.

Also, here's some more tidbits of knowledge I received from the author today:

just to emphasize, don't forget about the compressible nature of air. The *mass* flow of air (ie the lb/min) doesn't change from turbo inlet to turbo outlet to intake manifold. You can't destroy mass. The *volume* of that mass of air changes as it gets compressed by the turbo and then cooled by an intercooler. So the cfm goes down from turbo suction to discharge to intercooler outlet, while the lb/min stays the same for all 3 spots.

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Don't fall into the trap of "this turbo moves 500 cfm and that one moves 800 cfm, I'll get the bigger one so I'll flow more air." That just isn't true, the turbo will only flow what the engine can take. An 800 cfm turbo won't flow any more air on a given engine than the 500 cfm will, unless the 500 cfm one was undersized to begin with. It's like strapping on a 1100 cfm Holley Dominator on a mild small blcok chevy that needs a 650 cfm carb and expecting the engine to magically flow a bunch more air. If the engine can't take it, that extra flow capacity is worthless. Gotta match the turbo to the heads, cam, boost level, rpm, etc... just like you would match a carb to the engine's requirements.

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Again, I'm sure the wisemen and many others of you may know most or all of this, but to me it's still new, pretty cool stuff.

you can do 302.4(13.949)=X(press)
for these calcs. to work, you need to use the Kelvin scale 273K=0degC
and you are assuming it is an ideal gas

sorry, I meant
302.4/13.949=X/press
so 302.4/13.949=X/press is equal to 13.05
cooler temps=lower press.

also the 13.949 will change depending on boost level and restrictiveness of the intake ie filter mas...et cetera

well i know that 15lbs of boost on say a t25 is not the same as 15lbs of boost on a big16g or insert bigger turbo of your choice here, so I get confused there.

the lower boost pressure on a bigger turbo gives you more power because: more efficient turbo (assuming it is in its efficiency range) means cooler intake temps which means denser air which means more molecules to combine stoichiometrically with fuel molecules which means more power. The same pressure on a more efficient turbo gives you more power because you essentially have more air in a given volume.

One thing I had to think about is this....

Imagine you have a box that holds 550 CF. You can pack 550 cubic feet of air in it and it's nothing special. Or you can pack 1650 cubic feet in it and now you have boost (3psi if I am thinking straight). Sure you can do it, but your box will weigh more (mass) with 3x the air inside it. I think this can illustrate the difference in CFM and lb/min ratings. CFM is the physical size of your box. Lb/min is how much air you are cramming in it at a specifc rate.

Am I getting it or am I off base (as usual)?

Yeah, that makes sense...I don't know what I was thinking actually...I was thinking in terms of the formula and for some reason, it just didnt register that just because you have a given level of boost, it theory a bigger turbo will be flowing more air than a smaller one and hence will end up for more CFM's. It's been a long day

Originally posted by Violater101
well i know that 15lbs of boost on say a t25 is not the same as 15lbs of boost on a big16g or insert bigger turbo of your choice here, so I get confused there.

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Think of it this way, the engine at a given rpm draw a certain amount of air (this volume is fix and the formula is given on the page since we know the rpm and the capacity of the engine) so if the pressure in the system is constant at 15psi, how can one turbo flow more than the other? It doesnt. What's different is a t25 spin more rpm to push the air to maintain 15psi compared to a 16g. This makes the 16g compressor wheel more efficiency than the t25 because it pushing cooler air out. With these information (engine capacity, rpm, boost, temperature), you should be able to plot out a graph on the comp map of the turbo you wish to run and see how well it work.

Let's take for instance the T25 and a 16G both have the same TD05H wheel (which is not true but just for comparison purposes), then the only advantage the 16G has over the T25 is the better efficiency comp wheel. However, this is not the case, the 16G has a larger TW which allow more flow. So, remember the hot side plays an important role when picking a turbo because what you want is the most flow through the whole system.

I like this site....

http://www.smokemup.com/auto_math/airflow.php

Basically it does the equations shown above for you!

This page will calculate the mass of the air entering our engine. It is a good idea to play around with the number to get a feel for how temperature and engine Volumetric Efficiency affect the lbs/min of flow.

So lets enter the numbers for two different turbos.

T-25
1. Engine displacement is 122 cu inches for our engines.
2. Volumetric efficiency (VE) is approximately = 80% - 85% from what I have read. If you have a bigger turbine housing or ported intake/exhaust manifold the VE will go up. Lets use 80% because of a small turbine housing.
3. Boost at the manifold (lets try 15 psi).
4. RPM = 5700 (I choose this because my T-25 starts to die at this rpm)
5. Number of cylinders = 4
6. Air temperature entering your engine. Say 165 deg after intercooler.
This was just a guess. But can be calculated if you know the turbo effeciency intercooler efficiency and all the other issuses such as air temperature in and out Blah, Blah, Blah...

This will gives 20.86 lbs/min of air. From what I hear the t-25 will only flow about 20 to 22 lbs/min of air. So this sounds about right.

16G
1. 122 cu
2. 85% (more flow through the turbine housing)
3. 15 psi for comparison
4. 5700 rpm for comparison
5. say 120 deg

This gives 23.7 lbs/min of air.

So the 16g will flow 23.7/20.86 = approx. 15% more air at 15 psi and 5700 rpm. That is a pretty good gain!

The real benefit is that at higher rpms the 16g will still flow air, where as the t-25 will drop to about 10 psi because it cant keep up with the volumetric flow of the engine.

One important point--notice that when you keep everything the same but change the temperature the cfm stays the same; but the lbs/min changes!

Excellent post! I'm guessing that the reason you used 120 deg in the 16G calculation vs. 165 is because the bigger, more "efficient" wheel wouldn't heat up the intake charge as much as a T-25. Am I on the right track?

You are right; since the 16G is more efficient it will produce cooler charged air.

Those numbers were educated guesses because they depend on things like your intercooler efficiency, turbo efficiency, and outside temperature. They could be calculated except that there is not a turbo flow map for a T-25 available.

Good info. From what I can tell, that seems to be about right. I am guessing the DSM link's lbs/min ratio is fairly accurate. I was showing around 36lbs/min around 18lbs of boost around 6700rpms. I think the green is rated somewhere around 50 lbs/min, of course that won't happen unless I turn up the boost and have a better flowing head and intake.

Lbs/min change with temperature. What DSMlink used to calculate lbs/min base the temperature from your intake thermister (IAT) and not the actual temperature feeding into the engine. You have know what temperature actually going into the engine to get a close enough estimation.

Hi guys, at Doug's invitation I thought I would drop in and contribute if I may. I wrote the article on the turbo Buick site, and I'm glad I did since I've "met" (electronically anyway) lots of interesting people from all over the world who share our interests.

Just to let you know where I'm coming from, I'm not a automotive turbo expert. I am a chemical engineer that does design work in the refining and petrochemical industry. As such I have to deal with things like centrifugal compressors and turbines (among many other things). Biggest one I've done so far is a 4 stage compressor boosting from 5 psi to 600 psi that took a 30,000 hp turbine to drive it! I'm telling you this so you can judge whether or not to believe anything I say. Too many internet experts out there that might know what they are talking about, and then again they might not. I think I know these things, but my knowledge isn't specifically automotive in nature. The same principles apply though.

Anyway, on to work. DCJ98GST, I don't think I like the conclusion you arrived at with your example. I hope you don't mind me poking at it for a second. You came to the conclusion that a bigger turbo might flow 15% more at the same boost pressure. I think you made a bad assumption that led you to that answer, and that was the charge air temperature.

If you run through the math, you find that while an increase in compression efficiency does lower the compressor outlet temperature, its not that big of a gain.

The outlet temperature is only dependant on two things: the efficiency and the pressure ratio (or the outlet pressure/inlet pressure). And if we keep the boost level constant, we only have the efficiency to deal with.

At 15 psi boost the pressure ratio will be about 2.25. With a 60% compression efficiency and a 80F inlet temperature, the outlet temperature should be about 315F. Suppose you put some new turbo on and it has a compression efficiency of 70%. The new outlet temperature is about 280F. 35F drop is pretty good, but is it really worth? Turns out the colder outlet temp increases the air density by 4.5% If you had a nonintercooled car, that should mean (all else being equal) a 4.5% increase in mass air flow, and a 4.5% increase in power. Think about that, 10% better efficiency leads to a 35F drop in temperature, but that by itself is only worth 4.5% more air flow on a nonIC car.

But toss the IC into the equation. An air/air IC can at best get the air down to ambient temperature. If you study heat exchanger theory, you find that dropping the inlet temperature by 35F doesn't mean the outlet temperature also drops by 35F. The IC kind of dampens things, on the one hand it doesn't have as much heat to transfer with the colder inlet, but on the other hand the closer you get to ambient air temperature the harder it is to get there. Net effect, if you drop the inlet temperature by 35F then the outlet may only drop by 20F. And what is that worth? A 20F drop in the IC outlet only increases the charge air density (and therefore mass air flow and therefore power) by a little over 3%.

To sum all that up, on an intercooled car, an increase in compressor wheel efficiency from 60% to 70% (a pretty big jump) will only net a 3-4% gain in air flow *from temperature effects alone*.

When you changed the charge air temperature from 165 F to 120 F you cheated. That is why you got the big air flow increase. In reality it will be a lot less.

But still, people put on bigger turbos and do see more power at the same boost level. If its not air temperature, then what is it? I think it's two things (and this is my opinion, remember my qualifications, so take it for what its worth), and they are both on the exhaust side.

As you so correctly mentioned, when you swap turbos you swap turbine sides as well as compressor sides. A better flowing turbine side will reduce exhaust backpressure. Less backpressure means less exhaust left in the cylinder after the exhaust valve closes, which means more room for fresh air on the intake stroke. This means increased volumetric efficiency, or more air flow at the same charge air temperature and boost level. It's like adding more cubic inches! I believe that if you kept the compressor side exactly the same and changed the turbine side to something more freely flowing, you will see a power increase. And you do. Bigger turbine housings give a power increase, even though you haven't changed the compressor side at all. Might add some lag and such, but ultimately power is improved. It's similar in concept to getting rid of a cat converter or a restrictive muffler. Lower backpressure = improved VE = more air flow = more power. So thats one thing.

The other thing I think does come from improved compressor efficiency, but it isn't a temperature thing. It takes power to turn that compressor wheel, hp that is extracted from the exhaust by the turbine. If you improve the compression efficiency from 60% to 70% you need something like 20% less hp to drive the compressor wheel. Power to drive that wheel comes from exhaust temperature, exhaust flow, and exhaust pressure. If you reduce the power required by the compressor, then the exhaust flow that the turbine wheel needs goes down (ie more exhaust through the wastegate) and the exhaust pressure required goes down. And again, the lower backpressure means improved VE, more air flow, and more power.

Sorry this has been such a long book, but to sum up: IMO, a bigger turbo at the same boost level makes more power, but not due to temperature effects, but mostly due to improved VE. If you change compressor sides and don't change the turbine side, I don't think you'll see nearly the gains that someone who gets a whole new unit sees.


I could on, but I think I've done enough damage here for today

John

Net effect, if you drop the inlet temperature by 35F then the outlet may only drop by 20F. And what is that worth? A 20F drop in the IC outlet only increases the charge air density (and therefore mass air flow and therefore power) by a little over 3%.

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So would this also mean that all the efforts that some of us put into cold air intakes and air filter block off plates and the like in order to bring in a cool as possible intake charge is not really worth that much, HP-wise?

I think he was putting in perspective the ACTUAL changes. Taking a guestamate and putting real numbers to it. Remember a cooler charge also reduces detonation, which in turn makes more reliable output and less engine damage. This is why an intercooler is more necessary... kinda like water injection. Damn this is some good sh!t!!! I love it!! Sticky! Sticky!!

Doug - yeah, I didn't mean to imply that lower temperatures are worthless! Definitely a Good Thing! I was just trying to say that colder air temps alone don't explain why a bigger turbo makes more power. It sounds great, but when you run through the numbers a lower air temp just can't be the total explanation for the power gains that people sometimes claim with a bigger turbo at the same boost level. For that matter, plenty of Buick guys have swapped to larger turbos and not seen ANY power gains in the 15-17 psi range, but seen gains once they get up to the mid 20's. The reason being their old turbos were plenty efficient at the lower boost level, and their turbine sides didn't give a lot of backpressure there either, so there was no gain to be had. If they bought new turbos to run faster at the same boost level, they were wasting their money... Depends on what you have now and what are going to. Going from slightly too small to slightly too big might see a nice gain, while going from a little too big to way too big will probably net you nothing.

Anyway, cold air and better ICs, absolutely! Even if a lower temp on its own gives 1 or 2% more power, that's nothing to sneeze at. For that matter, you know the old saw from the car magazines that says a 10F drop in air temperature = 1% increase in power? This holds true for turbo cars as well, it's just that you have to look at the temperature in the intake manifold, not the outside air temp. If you check the charge air density at two temperatures 10F apart, you see that the density goes up about 1.5% (if I remember correctly) for that 10F drop. 1.5% increase in charge air density should give 1.5% more mass air flow, which should translate into 1-1.5% more power, give or take. Looks like a pretty good rule of thumb to me!

Even so, the mechanism of where the power comes from with these upgrades isn't always just the temperature.

Going from a stock air filter to a K&N for example, even at the same temperature provides a gain. Why? There is a suction pulled at the compressor inlet, if you put a vacuum gauge there you'll see a few inches of vacuum maybe. Anything you do to get that pressure closer to atmospheric pressure (ie less vacuum, 1" vac is better than 2" vac which is better than 4" vac) will give you power. Fix the boost pressure at some number, like 15 psi. The compression ratio for the turbo is P2/P1, or inlet pressure divided by outlet pressure (in absolute pressure of course). If you have a higher pressure at the inlet, then P2/P1 just got smaller. The turbo doesn't have to work as hard to make that 15 psi boost. The compressor outlet gets a little cooler (higher P2/P1 gives higher outlet temperatures). Less hp to turn the compressor wheel means lower exhaust backpressure, also a power gain. That's why free flowing, low pressure drop inlet systems are so important. Big pipes, as short as feasible, and big air filters are the order of the day for that reason.

A couple of more things that are more on topic:
if you know the cfm flow at some temperature and pressure, here's how to find out what the flow is at a different temperature and pressure:
V2 = V1 x (P1/P2)
and
V2 = V1 x (T2/T1)
and you can combine both and get
V2 = V1 x (P1 x T2)/(P2 x T1)

where V is volume flow, P is absolute pressure (gauge pressure + 14.7), and T is absolute temperature (deg F + 460)

suppose you know that 36 lb/min air = 520 cfm at 13.949 psia and 85 F. Your intake manifold is at 15 psi boost and 130 F. What is the cfm flow to the intake?

V1 = 520 cfm
T1 = 85F + 460 = 545 R
T2 = 130F + 460 = 590 R
P1 = 13.949 psia
P2 = 15 + 14.7 = 29.7 psia
V2 = 520 x (13.949 x 590)/(29.7 x 545) = 264.4 cfm.

So the 520 cfm at the compressor inlet has been reduced to 264.4 cfm at the intake manifold. You still have 36 lb/min of air flow at both spots of course.

Here's another one that helps answer that favorite question of bulletin boards everywhere: how much hp do I get from a pound of boost?

If you take (new boost pressure + 14.7), divide it by (old boost pressure + 14.7), and multiply the result by the hp made at the old boost level, you'll get the amount of hp made at the new boost level.

ie, Hp2 = Hp1 x (P2/P1)

The basic assumption implicit in the above is that an X% increase in charge air density gives the same X% increase in horsepower. I think this is a good estimate for small boost increases, but will give worse and worse answers as the difference between old and new boost level increases. I'd believe it up to 3 psi differences or so, but not much more than that, unless you're just looking at it as a guesstimate.

For accuracy, the following conditions have to be met:
1. the rpm stays the same
2. the air temperature to the intake manifold stays the same
3. the engine's volumetric efficiency and BSFC stay the same.

Example: go from 10 psi boost to 13 psi. The percent hp increase is approximately = (13+14.7)/(10+14.7) = 1.12, or a 12% increase.

So a 2.0L originally making 150 hp would now be making about 168 hp, an 18 hp gain (6 hp per psi), but 5.0L engine that was making 325 hp would now be making 365 hp, a 40 hp gain (13.3 hp per psi).

John

Originally posted by JDEstill
The other thing I think does come from improved compressor efficiency, but it isn't a temperature thing. It takes power to turn that compressor wheel, hp that is extracted from the exhaust by the turbine. If you improve the compression efficiency from 60% to 70% you need something like 20% less hp to drive the compressor wheel. Power to drive that wheel comes from exhaust temperature, exhaust flow, and exhaust pressure. If you reduce the power required by the compressor, then the exhaust flow that the turbine wheel needs goes down (ie more exhaust through the wastegate) and the exhaust pressure required goes down. And again, the lower backpressure means improved VE, more air flow, and more power.

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You are saying that the turbo's efficiency is relieving the engine from doing more work to turn the turbo.

One thing that contradicts what you are saying is lag.

We all can agree that at 15psi a 16G is more efficient than say the small t-25. But the T-25 will hit 15 psi at about 2700 rpm and the 16G maybe around 3400 rpm.

So in terms of the turbo using the exhaust flow's energy, it seems like the T-25 is more "efficient" to get to 15psi. This is due to the 16G having larger wheels with a larger polar moment of inertia. Some things like floating dual row ball bearing configurations can help a turbo spin faster due to less frictional energy wasted but that is another issue.

So it takes less power to turn the T-25 to 15 psi than the 16G.
Of course the 16G will help VE of the engine because of a larger turbing housing, but it will take more energy to turn the larger wheels to the same psi.

If a super efficient turbo with huge wheels take less energy to spin we would all be using them for street use! Because they would spool faster, but in reality some of these turbos take a tremendous amout of engine output to spin fast enough to start producing boost.

DCJ -

I have to admit that I'm not a mechanical engineer, just a humble process engineer. As such I am well equipped to talk steady state, but not so well equipped to discuss the dynamics of this particular process. So, spooling and lag I can only talk about in a general way.

I know that a larger turbine wheel and turbine housing will A. take longer to spool up, and B. allow more ultimate hp due to reduced backpressure. That's about all I can contribute there.

You said "So it takes less power to turn the T-25 to 15 psi than the 16G.
Of course the 16G will help VE of the engine because of a larger turbing housing, but it will take more energy to turn the larger wheels to the same psi."

I don't think that is quite correct, not to the way I think about things anyway. Steady state, a more efficient compressor wheel takes less hp to turn for a given set of conditions. You can't argue that. But that doesn't say anything about how it gets there. The fact that it takes longer to get there doesn't mean that it takes more hp to turn the wheel, it just means (to me anyway) that the turbine is less efficient at extracting hp from a given exhaust flow. That is the downside to a bigger turbine wheel and housing I believe.

If small turbine A gets 30 hp from some exhaust flow, and big turbine B gets 20 hp from that same exhaust flow, sure turbine A will spool faster, while turbine B offer more ultimate hp due to lower backpressure. But that does not mean that the compressor is taking more power to turn, it doesn't say anything about the compressor at all. And with a more efficient compressor wheel, there will be less backpressure required with either turbine.

Hope I got that thought clear...

Now if by "more energy required to turn the wheels to the same psi" you mean total amount of energy in the exhaust that is passing by the wheel, then I agree with you. Getting less power out of the exhaust that is there, even if you need less power to get the compressor wheel going, could wind up in a situation where the total energy required, defined as "energy extracted by turbine + available energy lost out the tailpipe", is higher with the bigger turbo, ie it would take more of that total to get things going. If that's what you meant, then I'm with you all the way.

Anyway, I agree that giant super efficient turbos aren't the way to go! There is the balancing act between transient response and ultimate power. Back in the late 80's one of the turbo upgrades for GN's had an oversized turbine housing. Proved to be worth 2 tenths in the quarter, but everyone hated it because it was a laggy SOB. People started swapping their old turbine housings onto the new turbo trying to fix that up.

John

Originally posted by JDEstill
Going from slightly too small to slightly too big might see a nice gain, while going from a little too big to way too big will probably net you nothing.

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Tons of good info, John, thanks-- you should write a book (well, you almost have!). Anyway, I think I've gotten a little (more) lost in the torrent of information being generated here, so I need to ask a basic question that may have already been covered, but my head is spinning. How do you calculate what size turbo is the right size for a particular application? I remember some stuff from the Summary paragraph on your web page about calculating volumetric flow and looking at compressor maps and things like that, so is that the best/only way to do it? Is there any way to at least get in the ballpark of desired turbo "size" (CFM or lbs/min or whatever) short of having to know how to read a compressor map? I mean, I know there's no magic answer where you just plug in a few numbers into a formula and it spits out the answer of which turbo is right for you, but if you could summarize again how to go about doing some research on making some informed turbo selections for our particular DSM applications it would help me (and others, I'm sure) a lot. Sorry if this is like asking someone, "Just tell me what it is that I need!"-- I didn't mean it that way.